Find the inverse of the matrix, $\text B = \left[\begin{array}{rr}-5 & 7 \\ -1 & 2\end{array}\right]$. Non-integers should be given either as decimals or as simplified fractions. $ B^{-1}=$
The Strategy To find the inverse of an invertible matrix, we can use Gaussian Elimination. To do this, we do the following. First, we append the matrix $\text B$ with the identity matrix $\text I$ to get [ B | I ] \left[\begin{array} ~\text B ~ |~\text I\end{array}\right]. Next, we use Gaussian Elimination to reduce $\text B$ to the identity matrix, $\text I$. Performing the same operations on $\text I$ will convert it to $\text B^{-1}$, so that our new matrix becomes [ I | B − 1 ] \left[\begin{array} ~\text I ~ |~\text B^{-1}\end{array}\right]. Appending $\text B$ with $\text I$ [ B | I ] = [ − 5 − 1 7 2 1 0 0 1 ] \left[\begin{array} ~\text B ~ |~\text I\end{array}\right]=\left[\begin{array}{rr}-5 & 7 & 1 & 0 \\ -1 & 2 & 0 & 1 \end{array}\right] Eliminating the leading term in the second row We want the first term of $R_2$ to equal $0$, so we subtract $\dfrac{1}{5}R_1$ from $R_2$. $\left[\begin{array}{rr}-5 & 7 & 1 & 0 \\ {-1} & {2} & {0} & {1} \end{array}\right]\xrightarrow{R_2-\dfrac{1}{5}R_1\rightarrow R_2}\left[\begin{array}{rr}-5 & 7 & 1 & 0 \\ {0} & {\dfrac{3}{5}} & {-\dfrac{1}{5}} & {1} \end{array}\right]$ Reducing the leading terms and back-solving Now, let's reduce the leading term of $R_2$ to equal $1$. $\left[\begin{array}{rr}-5 & 7 & 1 & 0 \\ {0} & {\dfrac{3}{5}} & {-\dfrac{1}{5}} & {1} \end{array}\right]\xrightarrow{\dfrac{5}{3}R_2\rightarrow R_2}\left[\begin{array}{rr}-5 & 7 & 1 & 0 \\ {0} & {1} & {-\dfrac{1}{3}} & {\dfrac{5}{3}} \end{array}\right]$ We are ready to back-solve to get [ I | B − 1 ] \left[\begin{array} ~\text I ~ |~\text B^{-1}\end{array}\right]. $\begin{aligned}\!\!\left[\begin{array}{rr}\!\!{-5}\! & \!\! {7}\! & {1}\! & {0} \\ \!0\! & 1\! & \!\! -\!\dfrac{1}{3}\! & \dfrac{5}{3} \!\end{array}\right]\!\xrightarrow{\!\!R_1\!-7R_2\!\rightarrow R_1\!\!} \!\!&\left[\begin{array}{rr}\!{-5}\!\! & {0}\!\! & \!\!\!{\dfrac{10}{3}} \!\!&\! {-\!\dfrac{35}{3}} \!\\ \!0\! & 1\! & -\!\dfrac{1}{3}\! & \dfrac{5}{3} \end{array}\right] \!\!\xrightarrow{\!\!-\!\dfrac{1}{5}\!\!R_1\!\rightarrow R_1\!\!}\!\!\left[\begin{array}{rr}{1}\!\! & {0}\!\! &\! {-\!\dfrac{2}{3}}\!\! & {\dfrac{7}{3}} \\ \!0\! & 1\! & -\!\dfrac{1}{3}\! & \dfrac{5}{3} \end{array}\right]\end{aligned}$ Therefore $\text B^{-1}=\left[\begin{array}{rr} -\dfrac{2}{3}\!\! & \dfrac{7}{3} \\ -\dfrac{1}{3} \!\!& \dfrac{5}{3} \end{array}\right]$. Summary $\text B^{-1}=\left[\begin{array}{rr} -\dfrac{2}{3}\!\! & \dfrac{7}{3} \\ -\dfrac{1}{3} \!\!& \dfrac{5}{3} \end{array}\right]$